I am stuck on the following problem:
Let $f:(a,b)\to \Bbb R$ be lower semicontinuous. Also $\lim_{x\to a+} f(x)=\lim_{x\to b-} f(x)=\infty$. Show that $\exists \rho\in (a,b)$ such that $f(\rho)\le f(x)$ forall $ x\in (a,b)$.
Since $f:(a,b)\to \Bbb R$ is lower semicontinuous so if we choose any point $x_0$ in the domain then given $\epsilon>0$ there exists $\delta>0$ such that $f(x_0)-\epsilon\le f(x) $ forall $x\in (x_0-\delta,x_0+\delta)$.
Also $\lim_{x\to a+} f(x)=\lim_{x\to b-} f(x)=\infty$ implies given any $M>0$ there exists $\delta_1,\delta_2>0$ such that $x\in (a,a+\delta_1)$, $x\in (b-\delta_2,b)$ implies $f(x)>M$ .
But I am unable to connect everything to obtain the required result.
What should I do? Can someone help?
Since $f$ is growing without boundary at the end points and you defined the function to map to the reals, $f$ needs to be bounded from below, that is to say $\inf_{(a,b)} f =: M > -\infty$. So we have to show that the infimum is actually attained at some point. For this take a sequence $(x_n \in (a,b))_{n\in\mathbb{N}}$ such that $$ \lim_{n\to \infty} f(x_n) = M. $$ Now, the sequence $(x_n)$ is bounded and so there is a convergent subsequence $x_{n_k}$ to $x$ (where $x$ cannot be either $a$ or $b$). So we get $$ f(x)\leq \lim\inf f(x_{n_{k}}) \leq \lim_{n\to \infty} f(x_{n_k}) = M, $$ from which we can read of $f(x) = M$ and hence $x = \rho$.
Hope that makes sense? Also notice that I used a different characterization for the lower semicontinuous via the $\lim \inf $. Are you familiar with it?